chemical formulas and compounds - nebula.wsimg.com

Chapter 7

Chemical Formulas and

Compounds

Chapter 7

Chemical Names and Formulas

– Covalent Formulas

Section 7.1 Part I

Chapter 7

Section 7.1.I Objectives

• Explain the significance of a chemical formula.

• Determine the formula of an ionic compound formed

between two given ions.

• Name an ionic compound given its formula.

• Using prefixes, name a binary molecular compound

from its formula.

• Write the formula of a binary molecular compound

given its name.

Chapter 7 Chapter 7 Section 1 Chemical Names and Formulas

Lesson Starter

• CCl4 MgCl2

• Use the formulas to guess the name of

these compounds

• What kind of information can you tell from

the formulas?

• Guess which compound is molecular and

which is ionic.

• Chemical formulas are the language of

chemistry and tell us a lot about the

substances they represent.

Chapter 7

Significance of a Chemical Formula

• A chemical formula indicates the relative number of

atoms of each kind in a chemical compound.

• For a Molecular compound - the chemical formula

shows the ACTUAL number of atoms of each element

in one molecule of the compound

• example: octane — C8H18

There are 8 C atoms in the

molecule.

There are 18 H atoms in

the molecule.

Section 1 Chemical Names and Formulas

Chapter 7 Chapter 7

Significance of a Chemical Formula, continued

• For an Ionic compound – the chemical formula

represents one formula unit —the simplest ratio of the

cations (+) and anions (-)

2

4SO

• Note also that there is no subscript for

sulfur: when there is no subscript next

to an atom, the subscript is understood

to be 1.


• aluminum sulfate — Al2(SO4)3

• Parentheses around the polyatomic ion

identify it as a unit. The subscript 3

refers to the unit.

Chapter 7 Chapter 7

Naming Binary Molecular Compounds

• Unlike ionic compounds, molecular compounds are

composed of individual covalently bonded units, or

molecules.

• LEAST electronegative element is listed first.

• We use prefixes to indicate the number of each type

of atom in a molecule.

• examples: CCl4 — carbon tetrachloride (tetra- = 4)

CO — carbon monoxide (mon- = 1)

CO2 — carbon dioxide (di- = 2)


Chapter 7

Prefixes for Naming Covalent Compounds

Chapter 7 Section 1 Chemical Names and Formulas

Chapter 7 Chapter 7

Naming Binary Molecular Compounds,

continued

Sample Problem D

a. Give the name for As2O5.

b. Write the formula for oxygen difluoride.


Chapter 7 Chapter 7


continued

Sample Problem D Solution

a. A molecule of the compound contains two arsenic

atoms, so the first word in the name is diarsenic.

The five oxygen atoms are indicated by adding the

prefix pent- to the word oxide.

The complete name is diarsenic pentoxide.


Chapter 7 Chapter 7


continued Sample Problem D Solution, continued

b. Oxygen is first because it is less electronegative

than fluorine.

Because there is no prefix, there must be only one

oxygen atom.

The prefix di- in difluoride shows that there are two

fluorine atoms in the molecule.

The formula is OF2.


Chapter 7 Chapter 7

Covalent-Network Compounds

• Some covalent compounds do not consist of

individual molecules.

• Instead, each atom is joined to all its neighbors in a

covalently bonded, three-dimensional network.

• Subscripts in a formula for covalent-network

compound indicate smallest whole-number ratios of

the atoms in the compound, just like in ionic

compounds.

• examples: SiC, silicon carbide

SiO2, silicon dioxide

Si3N4, trisilicon tetranitride.


Chapter 7 Chapter 7

Acids and Salts

• An acid is a certain type of molecular compound. Most

acids used in the laboratory are either binary acids or

oxyacids.

• Binary acids are acids that consist of two

elements, usually hydrogen and a halogen.

• Oxyacids are acids that contain hydrogen, oxygen,

and a third element (usually a nonmetal).


Chapter 7 Chapter 7

Acids and Salts, continued

• In the laboratory, the term acid usually refers to a solution in

water of an acid compound rather than the acid itself.

• example: hydrochloric acid refers to a water solution of the

molecular compound hydrogen chloride, HCl

• Many polyatomic ions are produced by the loss of hydrogen ions

from oxyacids.

• examples:

2

4SO

3NO

3

4PO

sulfuric acid H2SO4 sulfate

nitric acid HNO3 nitrate

phosphoric acid H3PO4 phosphate


Chapter 7 Chapter 7

Acids and Salts, continued

• An ionic compound composed of a cation and the anion from an acid is often referred to as a salt.

• examples:

• Table salt, NaCl, contains the anion from hydrochloric acid, HCl.

• Calcium sulfate, CaSO4, is a salt containing the anion from sulfuric acid, H2SO4.

• The bicarbonate ion, , comes from carbonic acid, H2CO3.

3HCO


Chapter 7

Chemical Names and Formulas

– Ionic Formulas

Section 7.1 Part II

Chapter 7

Section 7.1.II Objectives

• Explain the significance of a chemical formula.

• Determine the formula of an ionic compound formed

between two given ions.

• Name an ionic compound given its formula.

• Using prefixes, name a binary molecular compound

from its formula.

• Write the formula of a binary molecular compound

given its name.

Chapter 7 Chapter 7

Monatomic Ions

• Many main-group elements can lose or gain electrons to form ions.

• Monoatomic ions - ions formed from a single atom

• example: To gain a noble-gas electron configuration, oxygen gains two electrons to form O2– ions.

• Some main-group elements tend to form covalent bonds instead of forming ions.

• examples: carbon and silicon


Chapter 7 Chapter 7

Monatomic Ions, continued

Naming Monatomic Ions

• Monatomic cations - identified just by the element’s name

• examples:

• K+ = potassium cation

• Mg2+ = magnesium cation

• Monatomic anions - the ending of the element’s name is

dropped, and the ending -ide is added to the root name

• examples:

• F– = fluoride anion

• N3– = nitride anion


Chapter 7

Common Monatomic Ions


Chapter 7 Chapter 7

Binary Ionic Compounds

• Binary compounds - compounds composed of two elements

• In a binary ionic compound, total positive charge must equal total negative charge

• The formula for a binary ionic compound can be written given the identities of the compound’s ions.

• example: magnesium bromide

Ions combined: Mg2+, Br–, Br–

Chemical formula: MgBr2


Chapter 7 Chapter 7

Binary Ionic Compounds, continued

• To determine the formula for a binary ionic compound,

CROSS-OVER to balance charges between ions

• example: aluminum oxide

1) Write the symbols for the ions.

Al3+ O2–

3+ 2

2 3Al O

2) Cross over the charges by using the absolute

value of each ion’s charge as the subscript for the

other ion


Chapter 7 Chapter 7

Binary Ionic Compounds, continued

• example: aluminum oxide, continued

3+ 2

2 3Al O

3) Check the combined positive and negative

charges to see if they are equal.

(2 × 3+) + (3 × 2–) = 0

The correct formula is Al2O3


Chapter 7

Writing the Formula of an Ionic Compound


Chapter 7 Chapter 7

Naming Binary Ionic Compounds

• The nomenclature (naming system) for binary ionic compounds combines the names of the compound’s cations and anions.

• The name of the cation is given first, followed by the name of the anion:

• example: Al2O3 — aluminum oxide

• For most simple ionic compounds, the ratio of the ions is not given in the compound’s name, because it is understood based on the relative charges of the compound’s ions.


Chapter 7 Chapter 7

Naming Binary Ionic Compounds, continued

Sample Problem A

Write the formulas for the binary ionic compounds

formed between the following elements:

a. zinc and iodine

b. zinc and sulfur


Chapter 7 Chapter 7


Sample Problem A Solution

Write the symbols for the ions side by side. Write the

cation first.

a. Zn2+ I−

b. Zn2+ S2−

Cross over the charges to give subscripts.

a. 2+

1 2Zn I–

2+ 2

2 2Zn S –b.


Chapter 7 Chapter 7


Sample Problem A Solution, continued

Check the subscripts and divide them by their largest common factor to give the smallest possible whole-number ratio of ions.

a. Total charge of (1 × 2+) + (2 × 1−) = 0

The largest common factor of the subscripts is 1.

The smallest possible whole-number ratio of ions in the compound is 1:2.

The formula is ZnI2.


2+

1 2Zn I–

Chapter 7 Chapter 7


Sample Problem A Solution, continued

b. Total charge of (2 × 2+) + (2 × 2−) = 0

The largest common factor of the subscripts is 2.

The smallest whole-number ratio of ions in the

compound is 1:1.

The formula is ZnS.


2+ 2

2 2Zn S –

Chapter 7

The Stock System of Nomenclature

• Some elements like iron can form two or more cations

with different charges.

• To distinguish the ions formed by such elements,

scientists use the Stock system of nomenclature.

• The system uses a Roman numeral to indicate an ion’s

charge.

• examples: Fe2+ iron(II)

Fe3+ iron(III)

Chapter 7



Chapter 7 Chapter 7

Naming Binary Ionic Compounds, continued The Stock System of Nomenclature, continued

Sample Problem B

Write the formula and give the name for the compound

formed by the ions Cr3+ and F–.


Chapter 7 Chapter 7


Sample Problem B Solution

Write the symbols for the ions side by side. Write the

cation first.

Cr3+ F−


3+

1 3Cr F–


Chapter 7 Chapter 7


Sample Problem B Solution, continued

Total charge of (1 × 3+) + (3 × 1−) = 0

The largest common factor of the subscripts is 1, so the

smallest whole number ratio of the ions is 1:3.

The formula is

CrF3.


Chapter 7 Chapter 7


Sample Problem B Solution, continued

Chromium forms more than one ion, so the name of the

3+ chromium ion must be followed by a Roman numeral

indicating its charge.

The compound’s name is chromium(III) fluoride.


Chapter 7 Chapter 7

Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions

• Many common polyatomic ions are oxyanions—

polyatomic ions that contain oxygen.

• Some elements can combine with oxygen to form

more than one type of oxyanion.

• example: nitrogen can form or .

2NO

3NO

2NO

3NO

nitrate nitrite

• The name of the ion with the greater number of oxygen

atoms ends in -ate. The name of the ion with the smaller

number of oxygen atoms ends in -ite.


Chapter 7 Chapter 7

Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions, continued

• Some elements can form more than two types of

oxyanions.

• example: chlorine can form , , or . 3ClO

2ClO

4ClOClO

ClO

2ClO

3ClO

4ClO

• An anion with one LESS O atom than the -ite anion is

given the prefix hypo-.

• An anion with one MORE O atom than the -ate anion is

given the prefix per-.

hypochlorite chlorite chlorate perchlorate


Chapter 7

Polyatomic Ions


Chapter 7

Naming Compounds with Polyatomic Ions


Chapter 7 Chapter 7


Sample Problem C

Write the formula for tin(IV) sulfate.


Chapter 7 Chapter 7



4+ 2

4Sn SO –

4+ 2

2 4 4Sn (SO ) –

Add parentheses around the polyatomic ion if

necessary.


Chapter 7 Chapter 7


Sample Problem C Solution, continued

The total positive charge is 2 × 4+ = 8+.

The total negative charge is 4 × 2− = 8−.

The largest common factor of the subscripts is 2, so the

smallest whole-number ratio of ions in the compound is

1:2.

The correct formula is therefore Sn(SO4)2.


Chapter 7

Using Chemical Formulas

Section 7.3

Chapter 7

Section 7.3 Objectives

• Calculate the formula mass or molar mass of any

given compound.

• Use molar mass to convert between mass in grams

and amount in moles of a chemical compound.

• Calculate the number of molecules, formula units, or

ions in a given molar amount of a chemical

compound.

• Calculate the percentage composition of a given

chemical compound.

Chapter 7 Chapter 7

• A chemical formula indicates:

• the elements present in a compound

• the relative number of atoms or ions of each

element present in a compound

• Chemical formulas also allow chemists to calculate a

number of other characteristic values for a compound:

• formula mass

• molar mass

• percentage composition

Section 3 Using Chemical Formulas

Chapter 7 Chapter 7

Formula Masses

• The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula.

• example: formula mass of water, H2O average atomic mass of H: 1.01 amu average atomic mass of O: 16.00 amu


1.01 amu2 H atoms 2.02 amu

H atom

16.00 amu1 O atom 16.00 amu

O atom

average mass of H2O molecule: 18.02 amu

Chapter 7 Chapter 7

Formula Masses

• The mass of a water molecule can be referred to as a

molecular mass.

• The mass of one formula unit of an ionic compound,

such as NaCl, is not a molecular mass.

• The mass of any unit represented by a chemical

formula (H2O, NaCl) can be referred to as the formula

mass.


Chapter 7 Chapter 7

Formula Masses, continued

Sample Problem F

Find the formula mass of potassium chlorate, KClO3.


Chapter 7

Formula Masses, continued

Sample Problem F Solution, continued

Chapter 7

39.10 amu1 K atom 39.10 amu

K atom

35.45 amu1 Cl atom 35.45 amu

Cl atom

16.00 amu3 O atoms 48.00 amu

O atom


formula mass of KClO3 = 122.55 amu

Chapter 7 Chapter 7

Molar Masses

• The molar mass of a substance is equal to the mass

in grams of one mole, or approximately 6.022 × 1023

particles, of the substance.

• example: the molar mass of pure calcium, Ca, is 40.08 g/mol

because one mole of calcium atoms has a mass of 40.08 g.

• The molar mass of a compound is numerically equal

to the formula or molecular mass, but has a different

unit.


Chapter 7

Calculating Molar Masses for Ionic Compounds

Chapter 7 Section 3 Using Chemical Formulas

Chapter 7 Chapter 7

Molar Masses, continued

Sample Problem G

What is the molar mass of barium nitrate, Ba(NO3)2?


Chapter 7

Molar Masses, continued

Sample Problem G Solution

One mole of barium nitrate, contains one mole of Ba, two moles of N (1 × 2), and six moles of O (3 × 2).

Chapter 7

137.33 g H1 mol Ba 137.33 g Ba

mol Ba

14.01 g2 mol N 28.02 g N

mol N

16.00 g O6 mol O 96.00 g O

mol O


molar mass of Ba(NO3)2 = 261.35 g/mol

Chapter 7 Chapter 7

Molar Mass as a Conversion Factor

• RECALL: The molar mass of a compound can be

used as a conversion factor to relate an amount in

moles to a mass in grams for a given substance.


Chapter 7 Chapter 7

Molar Mass as a Conversion Factor, continued

Sample Problem H

What is the mass in grams of 2.50 mol of oxygen gas?


Chapter 7 Chapter 7


Sample Problem H Solution, continued

Calculate the molar mass of O2.

16.00 g O2 mol O 32.00 g

mol O


Use the molar mass of O2 to convert moles to mass.

2.50 mol O2 =

80.0 g O2

32.00 g O2

1 mol O2

Chapter 7 Chapter 7


Sample Problem I

Ibuprofen, C13H18O2, is the active ingredient in many

nonprescription pain relievers. Its molar mass is

206.31 g/mol.

a. If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle?

b. How many molecules of ibuprofen are in the bottle?

c. What is the total mass in grams of carbon in 33 g of ibuprofen?


Chapter 7 Chapter 7


Sample Problem I Solution, continued




206.31 g/mol.

a. If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle?

33 g C13H18O2 =

206.31 g C13H18O2

0.16 mol C13H18O2

1 mol C13H18O2

Chapter 7 Chapter 7






206.31 g/mol.

b. How many molecules of ibuprofen are in the bottle?

0.16 mol C13H18O2

= 1 mol C13H18O2

9.6x1022 molecules C13H18O2

6.022x1023 molecules

Chapter 7 Chapter 7






206.31 g/mol.

c. What is the total mass in grams of carbon in 33 g of ibuprofen?

0.16 mol C13H18O2 =

25 g C 1 mol C 1 mol C13H18O2

13 mol C 12.011 g C

Chapter 7 Chapter 7

Percentage Composition

• It is often useful to know the percentage by mass of

a particular element in a chemical compound.


• The mass percentage of an element in a compound

is the same regardless of the sample’s size.

𝒎𝒂𝒔𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝟏 𝒎𝒐𝒍 𝒐𝒇 𝒂 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅

𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒂 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 ∗ 𝟏𝟎𝟎 = % 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅

• The percentage by mass of each element in a

compound is known as the percentage

composition of the compound.

Chapter 7

Percentage Composition of Iron Oxides

Chapter 7 Section 3 Using Chemical Formulas

Chapter 7 Chapter 7

Percentage Composition, continued

Sample Problem J

Find the percentage composition of copper(I) sulfide,

Cu2S.


63.55 g Cu2 mol Cu 127.1 g Cu

mol Cu

32.07 g S1 mol S 32.07 g S

mol S

Molar mass of Cu2S = 159.2 g

Chapter 7 Chapter 7

Percentage Composition, continued

Sample Problem J Solution, continued

2

127.1 g Cu100

159.2 g C79.8 u

S5% C

u

2

32.07 g S100

159.2 g C20. S

u S15%


Chapter 7

Determining Chemical

Formulas

Section 7.4

Chapter 7

Section 7.4 Objectives

• Define empirical formula, and explain how the term

applies to ionic and molecular compounds.

• Determine an empirical formula from either a

percentage or a mass composition.

• Explain the relationship between the empirical formula

and the molecular formula of a given compound.

• Determine a molecular formula from an empirical

formula.

Chapter 7 Chapter 7 Section 4 Determining Chemical Formulas

Lesson Starter

• Compare and contrast models of the molecules NO2

and N2O4.

• The numbers of atoms in the molecules differ, but the

ratio of N atoms to O atoms for each molecule is the

same.

Chapter 7

Empirical and

Actual Formulas

Chapter 7 Section 4 Determining Chemical Formulas

Chapter 7 Chapter 7

• An empirical formula consists of the symbols for the

elements combined in a compound, with subscripts

showing the smallest whole-number mole ratio of the

different atoms in the compound.

• For an ionic compound, the formula unit is usually the

compound’s empirical formula.

• For a molecular compound, however, the empirical

formula does not necessarily indicate the actual

numbers of atoms present in each molecule.

• example: the empirical formula of the gas diborane is BH3,

but the molecular formula is B2H6.

Section 4 Determining Chemical Formulas

Chapter 7 Chapter 7

Calculation of Empirical Formulas

• To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition.

• Assume that you have a 100.0 g sample of the compound.

• Then calculate the amount of each element in the sample.

• example: diborane • The percentage composition is 78.1% B and 21.9% H.

• Therefore, 100.0 g of diborane contains 78.1 g of B and 21.9 g of H.


Chapter 7 Chapter 7

Calculation of Empirical Formulas, continued

• Next, the mass composition of each element is

converted to a composition in moles by dividing by the

appropriate molar mass.

1 mol B78.1 g B 7.22 mol B

10.81 g B

1 mol H21.9 g H 21.7 mol H

1.01 g H

• These values give a mole ratio of 7.22 mol B to

21.7 mol H.


Chapter 7 Chapter 7


• To find the smallest whole number ratio, divide each

number of moles by the smallest number in the

existing ratio.

7.22 mol B 21.7 mol H: 1 mol B : 3.01 mol H

7.22 7.22

• Because of rounding or experimental error, a

compound’s mole ratio sometimes consists of

numbers close to whole numbers instead of exact

whole numbers.

• In this case, the differences from whole numbers may be

ignored and the nearest whole number taken.


Chapter 7 Chapter 7


Sample Problem L

Quantitative analysis shows that a compound contains

32.38% sodium, 22.65% sulfur, and 44.99% oxygen.

Find the empirical formula of this compound.


Chapter 7 Chapter 7


Sample Problem L Solution, continued

1 mol Na32.38 g Na 1.408 mol Na

22.99 g Na

1 mol S22.65 g S 0.7063 mol S

32.07 g S

1 mol O44.99 g O 2.812 mol O

16.00 g O


Chapter 7 Chapter 7


Sample Problem L Solution, continued

Smallest whole-number mole ratio of atoms: The compound contains atoms in the ratio 1.408 mol Na:0.7063 mol S:2.812 mol O.

1.408 mol Na 0.7063 mol S 2.812 mol O: :

0.7063 0.7063 0.7063

1.993 mol Na :1 mol S : 3.981 mol O

Rounding yields a mole ratio of 2 mol Na:1 mol S:4 mol O.

The empirical formula of the compound is Na2SO4.


Chapter 7 Chapter 7

Calculation of Molecular Formulas

• The empirical formula contains the smallest possible whole numbers that describe the atomic ratio.

• The molecular formula is the actual formula of a molecular compound.

• An empirical formula may or may not be a correct molecular formula.

• The relationship between a compound’s empirical formula and its molecular formula can be written as follows.

x(empirical formula) = molecular formula


Chapter 7 Chapter 7

Calculation of Molecular Formulas, continued

• The formula masses have a similar relationship.

x(empirical formula mass) = molecular formula mass

• To determine the molecular formula of a compound, you must know the compound’s formula mass.

• Dividing the experimental formula mass by the empirical formula mass gives the value of x.

• A compound’s molecular formula mass is numerically equal to its molar mass, so a compound’s molecular formula can also be found given the compound’s empirical formula and its molar mass.


Chapter 7

Comparing Empirical and Molecular Formulas

Chapter 7 Section 4 Determining Chemical Formulas

Chapter 7 Chapter 7


Sample Problem N

In Sample Problem M, the empirical formula of a

compound of phosphorus and oxygen was found to be

P2O5. Experimentation shows that the molar mass of

this compound is 283.89 g/mol. What is the

compound’s molecular formula?


Chapter 7 Chapter 7


Sample Problem N Solution

Given: empirical formula

Unknown: molecular formula

Solution:

x(empirical formula) = molecular formula

x

molecular formula mass

empirical formula mass


Chapter 7 Chapter 7


Sample Problem N Solution, continued

Molecular formula mass is numerically equal to molar mass.

molecular molar mass = 283.89 g/mol

molecular formula mass = 283.89 amu

empirical formula mass

mass of phosphorus atom = 30.97 amu

mass of oxygen atom = 16.00 amu

empirical formula mass of P2O5 =

2 × 30.97 amu + 5 × 16.00 amu = 141.94 amu


Chapter 7 Chapter 7


Sample Problem N Solution, continued

Dividing the experimental formula mass by the

empirical formula mass gives the value of x.

x =

283.89 amu2.0001

141.94 amu

The compound’s molecular formula is therefore P4O10.

2 × (P2O5) = P4O10


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